anonymous
  • anonymous
Hello everyone. It's been awhile since I've been up here. I have one about a polynomial / another polynomial. Here it is: (x^2-y^2+2yz-z^2)/(x+y-z) Any thoughts?
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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whpalmer4
  • whpalmer4
You do long division on polynomials just like on numbers. What is the biggest thing you could multiply (x+y-z) by and have the result be < or = to the numerator? After you subtract the result from the numerator, what is left? Is there another value you could multiply (x+y-z) by that would be less than or equal to that? If so, that's another term in your answer, otherwise, what's left is the remainder, if it isn't 0.
anonymous
  • anonymous
It's a long division problem. The book answer is x-y+z. But how do you get there doing the long division?
anonymous
  • anonymous
It would be pretty easy if it was just a poly / a monomial. But this thing was tricky to me.

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anonymous
  • anonymous
Also I don't see how you could subtract the give denominator from the given numerator when there aren't any like terms on one versus the other.
anonymous
  • anonymous
*given
anonymous
  • anonymous
A little easier said that done huh?
whpalmer4
  • whpalmer4
You gotta have faith :-) \[(x^2 - y^2 + 2yz - z^2) / (x + y - z)\] Okay, by looking it, we know our answer has to have an x, a y, and a z, right? That's the only way we'll get x^2, y^2 and z^2 terms. We'll do this as a "speculative division" — we'll speculate about the answer, and then see if works. What we do will also bear some similarity to solving multiple equations with combination. Let's make a little table of what {x, y, z} * (x+y-z) gives us: \[x(x+y-z) = x^2 + xy - xz\]\[y(x+y-z) = y^2 + xy - yz\]\[z(x+y-z) = -z^2 + xz + yz\] Those are our building blocks. How do we combine them to get our original numerator? Well, we need an x in the quotient to produce the x^2 term. We need a -y in the quotient to produce the -y^2 term. We need a z in the quotient to produce the -z^2 term. Let's just add up those equations and see if we get the right thing! \[x(x+y-z) - y(x+y-z) + z(x+y-z)=\]\[x^2+xy-xz - (y^2+xy-yz) + (-z^2+xz+yz)=\] \[x^2 + xy - xz -y^2 -xy -(-yz) -z^2 +xz +yz=\]\[x^2-y^2 -z^2 + xy -xy +xz- xz +yz+yz \] and after the dust settles and we cross off all the terms that combine to 0, we have\[x^2-y^2+2yz -z^2\] which is our original numerator. Therefore, the quotient we "guessed", (x - y + z), is correct!
anonymous
  • anonymous
It took me a few minutes. Let me look at what you have there. Thanks for the help too.
anonymous
  • anonymous
I will have to test that theory on other problems, but thanks for that. I really appreciate it. However, do you know how to do it using the long division?
anonymous
  • anonymous
In other words, suppose I don't have a guess. Then what?
whpalmer4
  • whpalmer4
Well, it really *is* what you do when you do long division, I just organized the work a little differently.
anonymous
  • anonymous
Oh ok. I got you.
anonymous
  • anonymous
I will try that over the weekend on some other problems.
whpalmer4
  • whpalmer4
x - y + z x+y-z x^2 - y^2 + 2yz - z^2 x^2 xy -xz --------------------------------------------------- -y^2 + 2yz - z^2 - xy +xz -y^2 +yz - xy -------------------------------------------- yz - z^2 +xz yz -z^2 +xz ---------------------------------
whpalmer4
  • whpalmer4
I tend to make mistakes when doing lots of algebra, so if I can think a little first and simplify the pencil-pushing, my accuracy goes up :-)
whpalmer4
  • whpalmer4
Writing it out like a long division problem, it took me a couple of tries before I avoided making any mistakes. Doing it the "combination" way, I did it the first time, and it took less time, too!
anonymous
  • anonymous
That was the first time I've seen it done that way. I am going to look that up on youtube later.
whpalmer4
  • whpalmer4
I agree it's pretty weird seeing all of these "other" terms popping up as you do the long division, but like I said, you just have to have faith that they will all go away in the end :-)
anonymous
  • anonymous
It kind of threw me for a loop, but I will have this figured out by tomorrow.
whpalmer4
  • whpalmer4
That's the attitude I like to hear! :-) Too many people just say "oh, this is hard" or "I'm dumb" or whatever. Much of this stuff isn't really hard at all — it just takes the right explanation, and practice.
anonymous
  • anonymous
Well, thanks again.
anonymous
  • anonymous
I'm a fan now. Ttyl.
anonymous
  • anonymous
Use PEMDAS!

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